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Integration in the Complex Plane
Tables and Figures Seminar 16:01 ok then 16:02 so in the last seminar we covered differentiation of functions C -> C (where C denotes the complex numbers) 16:02 generalizing the notion of various limits in doing so 16:02 there are mainly two things you should have taken from that seminar: 16:03 1. many things that you know about real analysis continue to hold in the complex case. These include the familar rules of differentiation etc 16:04 2. a complex function to be differentiable imposes a certain rigidity. This is expressed most explicitly in the Cauchy-Riemann equations (CR) 16:04 We will now consider consider integration 16:05 Recall that in R, an integral is defined by the function we integrate and the interval of integration, where we have (for sufficiently nice functions) an intuitive geometric interpretation as an (oriented) area 16:06 now the straigtforward generasization of a real interval is a curve in C, where I don't want to define in detail what a curve is (because the details are somewhat messy) 16:06 but one thing is clear: if we have a sufficiently nice function gamma:a;b -> C then its image is a curve 16:06 (where a and b are real) 16:07 there is one problem in generalizing the integral to arbitrary curves, and that is that there is no directly intuitive equivalent to "area" 16:08 however, we proceed in formal analogy to the real case, and just see what we get (it will be different from the contour integral in R^2) 16:09 consider figure 3 on page two 16:09 I will try to convey the geometric idea of what we mean by an integral of f(z) "along the curve Gamma" 16:09 the (smooth) arc in the figure is supposed to represent a curve 16:10 and the three line segments represent a polygonal approximation to that curve 16:11 I call them d_1, d_2 and d_3, where d_i is to be taken as the *complex* vector that represents the line segment 16:12 now arbitrarily on the line segment represented by d_i we chose a complex number c_i (so c_i = d_1 + d_2 + ... +d_{i-1} + r*d_i, where r is in 1) 16:12 and then we define as an approximation of the integral (i.e. a riemann sum) d_1 * f(c_1) + d_2 o f(c_2) + d_3 * f(c_3) 16:13 ups, the o should of course be a * 16:13 now it is clear that there are infinitely many ways of approximating the curve, and infinitely many ways to choose the c_i 16:14 but for any choice of c_i and d_i we define I_{alpha} = sum_{i} f(c_i) * d_i 16:15 (where alpha is supposed to label the choice of c_i and d_i) 16:16 and then if, regardless of how we choose the d_i, so long as they approximate the curve linearly, and regardless of how we choose the c_i, so long as they lie on the d_i, the I_{alpha} converges to the same value if we use increasingly better approximations (increasingly many n), then we say that the common limit is the integral of f along Gamma 16:17 now this is a rather messy and long wordy description, but we will see in a minute that there is in fact an easy expression for that in terms of things we know 16:18 but before proceeding along this line, let me demonstrate that we can actually use this geometric definition to evaluate integrals 16:18 suppose we want to calculate the integral of f(z) = 1/z along the unit circle (oriented counterclockwise), and suppose we know that the integral exists 16:19 now consider figure 4 16:19 we approximate the unit circle by tangents 16:19 we will choose all of the line segments to have a common length d 16:20 we first need to find an expression of d_i 16:21 let's say that d_i is the tangent segment of length d at z_i, where z_i is somewhere on the unit circle 16:22 then, by definition, d_i is orthogonal to z_i, i.e. d_i is proportional to i*z_i (recall that multiplication by i means rotation by 90 degrees). But the length of z_i is unity, so d_i = i*z_i*d 16:23 next we choose c_i = z_i, then f(c_i) = f(z_i) = 1/z_i = conj(z_i), as |z_i|=1, where conj(z) denotes the complex conjugate of z. 16:24 thus in our riemann sum approximating the integral, the ith term will equal d_i*f(c_i) = d*z_i*i*conj(z_i) = d*i*|z_i| = d*i, as we chose z_i on the unit circle 16:25 so we can pull a factor of i out of the sum, and are left with sum_{i} d. But in the limit as we get better and better approximations, this sum must clearly equal the length of our curve 16:25 so we have shown that the integral of 1/z along the unit circle = 2pi*i 16:26 this is a very fundamental integral that occurs numerous times in complex analysis 16:26 but let us now return to a more formal definition of the contour integral 16:27 you can convince yourself that, if we parametrize a curve Gamma as gamma:a;b -> C, the wordy description from above reduces to integral_a^b f(gamma(t)) * gamma'(t) dt 16:28 in fact, if you dislike the geometric description, we can take this as the essential definition of the contour integral 16:29 for one thing, it can be seen easily that this definiton is well-behaved in the sense that the integral does not depend on the parametrization of Gamma. this can be shown most easily by using the substitution rule for functions R->C which can proved easily 16:30 but let us instead redo the above integral using the parametric evaluation 16:31 we can of course parametrize the unit circle by gamma(t) = e^{it}, for t in 2pi. Then gamma'(t) = i*e^{it}, so I = integral_0^{2pi} 1/e^{it} * i * e^{it} dt = 2pi*i 16:32 we have indeed proved more, for in the geometric evaluation we have assumed that the integral exists, which we didn't need here 16:32 so much for the definition 16:32 I have spent quite a lot of time on this because I think it is important to have a geometric view of the things 16:33 let us look at some basic properties of the integral defined this way 16:33 all of them can be shown easily using the parametric evaluation 16:34 first, the integral is linear in the integrand: integral_{Gamma} + b*g(z) dz = a*integral_{Gamma} f(z) dz + b*integral_{Gamma} g(z) dz 16:34 but furthermore, the integral is also "linear in the contour" 16:35 for if we denote by Gamma = Gamma_1 + Gamma_2 the contour (path) that first traverses Gamma_1, then Gamma_2, and by -Gamma the path that traverses in the opposite direction, the following holds 16:36 integral_{Gamma_1 + Gamma_2} f(z) dz = integral_{Gamma_1} f(z) dz + integral_{Gamma_2} f(z) dz 16:36 and integral_{-Gamma} f(z) dz = -integral_{Gamma} f(z) dz 16:36 all of these are trivial 16:37 now let's have a look at some of the more interesting properties 16:37 I stressed in the beginning that a function being complex differentiable imposed a large certain amount of rigidity, and we can also see this in integration, for the following holds: 16:38 if f(z) is complex differentiable inside Gamma, then the loop integral of f around Gamma vanishes 16:39 this is were the aforementioned difficulties with the definition of a curve surface again (what is inside and what outside), and I won't go into details here. the keywords are jordan curve, piecewise smooth curve, and so on 16:40 I won't prove this theorem either, here is a rough sketch: you can write integral_{Gamma} f(z) dz = integral_{Gamma} + iv(x,y) (dx + idy), then use Green's theorem to convert that into a surface integral, and simplify using the CR 16:41 it was of course obvious from the start that the CR played an essential role here 16:41 you might think (if you haven't seen comparable things before) that vanishing loop integrals are not very interesting, but something very remarkable follows from this: 16:42 if f(z) is complex differentiable inside an open, simply connected set G, then integrals along contours inside that set only depend on the beginning and end points, i.e. they are path independent 16:43 consider figure 5 16:43 let's assume that our function is differentiable on the entire sheet. Then what I just said means that the integral along A is the same as the integral along B 16:44 from that figure, we can also easily see the relation to vanishing loop integrals: 16:44 from the theorem, we know that (symbolically) integral_{A+B} = 0 = integral_{A+C}. 16:45 But from the basic properties, integral_{A+B} = integral_A + integral_B, so we have integral_A + integral_B = integral_A + integral_C, which of course implies integral_A = integral_B (make sure you see that geometrically) 16:45 or 16:45 er 16:46 the last statement should of course read integral_B = integral_C 16:47 so that means that we can continuosly deform a contour of integration in any way we like, so long as we stay in the region of holomorphy 16:47 (holomorphic is another term for complex differentiable) 16:48 so the integral from above immediatly tells us that the integral around *any* counterclockwise closed loop of f(z) = 1/z equals 2pi*i 16:49 here we of course assume that the loop goes aronud the origin, but doesn't "wind" around it more that once either 16:49 we also know that the loop integral around any other loop (not enclosing the origin) must be zero 16:50 there are a number of other interesting results on contour integration with closed loops, treating cases where there are singularities inside the contour, i.e. where the former theorem does not apply 16:51 we can begin with a general idea: 16:51 suppose we have a complicated loop, as depicted in the upper left image of figure 6 16:52 the function is assumed to be complex differentiable everywhere, except for the dots (we call them isolated singularities) 16:52 we can now try to continuously deform the contour of integration, without passing over the poles of course 16:52 our findings on path independence ensure that we don't change the value of the integral along the contour 16:53 I have tried to indicate one especially useful deformation in the other two images of figure 6 16:53 you see that we can reduce the complicated curve to loop integrals around the singularities 16:56 and you can see that the value of the integral essentially only depends on the number of times that the curve "winds" around each singularity 16:56 We can define Ind 16:56 argh 16:57 We can define Ind_{Gamma} z to be "the number of times Gamma winds around z" 16:57 this of course must be an integer 16:57 we count windings in the counterclockwise orientation positive, clockwise orientation counts negative 16:58 we can further define Res_a f to be 1/{2pi*i} lim_{r -> 0} int_{C_r} f(z) dz, where C_r is a circle of radius r and center a, oriented counterclockwise 16:59 the factor of 1/{2pi*i} might look useless but it turns out to be useful ;) 17:00 then, for any Gamma inside a set G where f is holomorphic except for a finite number of isolated singularities, we know that integral_{Gamma} f(z) dz = 2pi*i sum_{a in G} Ind_{Gamma} (a) Res_a f. 17:01 The sum reduces to a finite sum over the the singularities 17:01 now time is over, so let me just say a few more words: 17:02 it turns out that for sufficiently nice functions (those which are a ratio of two holomorphic functions, also called meromorphic functions) there is a relatively easy, different way to compute Res_a f 17:03 If we formalize these ideas, we get the Residue Theorem, which is usually attributed to cauchy 17:03 Res_a f is called the residue of f at a 17:04 if you look things up (you are prepared for this now), you will see that the residue can be computed by differentiation! 17:04 so this is one really nice method to evaluate integrals 17:04 there is another theorem (perhaps more fundamental) about loop integration, called the cauchy formula 17:05 it can be used to show that complex differentiable functions are necessarily smooth and analytic 17:05 so you have a great unity in complex analysis, conformal = differentiable = smooth = analytic, and we use the term holomorphic becaus we don't like the others ;) 17:06 complex analysis finds numerous applications 17:06 and a lot more can be said about it, especially when striving for formal proofs 17:06 thanks for listening 17:07 I hop I didn't bore everyone away ^^ 17:07 are there questions? 17:08 I think it is somewhat typical for these to go longer than an hour, or at least not uncommon 17:08 hm, the ones I saw where about one hour 17:08 <_llll_> think it's varied a fair bit 17:08 its entirely up to the presenter, thank you ness 17:09 i did only catch parts of this, but this wasn't very rigorous. for example the "definition" of the index 17:09 sure 17:09 it's asking a bit much to go through all this rigorously; it would fill several weeks in a complex analysis class 17:09 I presented it in a way I would present it to someone who hasen't done any more difficult analysis than real calculus 17:10 <_llll_> plenty of books do things completely rigorously, i dont know what is usually recommeded here 17:11 if there is interest, I could give one more seminar, fleshing out calculation of residues, the cauchy formula, perhaps liouville's theorem as an example of how these things can be applied. But the important stuff I wanted to say has been said. 17:11 I think it's definitely worth going through laurent series and computing residues 17:12 sure 17:21 whoop, i walked away, but i agree 17:29 * ness (n=ness@xx.xxx.xxx.xxx) Quit (Remote closed the connection) Category:Seminar Category:Complex Analysis